Theorem: For any two sets $A$ and $B$ in $\mathbb{R}^n$, we have $$m^{*}(A \cup B) \le m^{*}(A) + m^{*}(B).$$
Moreover, I can show that there exist two disjoint sets $A$ and $B$ such that $$m^{*}(A \cup B) < m^{*}(A) + m^{*}(B)$$ by using the contradiction to show that every set in $\mathbb{R}^n$ is measurable (this is unreasonable since the Vitali set is a non-measurable set in $\mathbb{R}^n$)
My question is whether we can find the form of any two sets like this? The form of them is trivial?
We know outer measure restricted to Lebesgue sigma algebra is countably additive.
If $A, B$ are disjoint measurable sets then $m^*(A\cup B) =m^*(A) +m^*(B) $
Hence for disjoint measurable sets strictly sub-aditivity of Lebesgue outer measure is not possible.
To get strictly sub-aditivity of outer measure choose $B\subset [0, 1]$ a Bernstein set .
Then Bernstein sets are non measurable and has full outer measure. Complement of a Bernstein is also Bernstein.
Then
$m^*(B\cup B^c)=m^*([0,1])=1<2=m^*(B) +m^*(B^c) $